Answer:
Step-by-step explanation:
In ΔMOT and ΔMPR,
OT║PR and MR is a transverse,
Therefore, ∠MTO ≅ ∠MRP [Corresponding angles]
Similarly, OT║PR and MP is a transverse,
Therefore, ∠MOT ≅ ∠MPR [Corresponding angles]
ΔMOT ~ ΔMPR → [By AA postulate of similarity]
[tex]\frac{MP}{MO}= \frac{MR}{MT}[/tex] [Reason → Corresponding sides of the similar triangles are proportional]
[tex]\frac{MO+OP}{MO}=\frac{MT+TR}{MT}[/tex] [By segment addition postulate]
1 + [tex]\frac{OP}{MO}[/tex] = 1 + [tex]\frac{TR}{MT}[/tex]
[tex]\frac{OP}{MO}=\frac{TR}{MT}[/tex]
In ΔTSR,
∠1 ≅ ∠2 [Given]
Therefore, TR ≅ SR [Sides opposite to the equal angles in a triangle are equal]
Therefore, [tex]\frac{OP}{MO}=\frac{SR}{MT}[/tex]
Or [tex]\frac{MO}{OP}=\frac{MT}{SR}[/tex]
Hence proved.
